\section{The parallel plane transmission line model of EM propagation in free space}
\label{sec:prob1}

\subsection{About waves and waveguides} 

Electromagnetic plane waves propagating in an unbounded medium have components of neither electric nor magnetic fields along their propagation direction.
The wave's electric and magnetic fields oscillate in the plane transverse to the direction of propagation, and are therefore called transverse waves. 
As far as our empirical experience tells us, electromagnetic waves can propagate throughout mediums other than the free space. When waves propagate through bounded 
regions, they are referred as {\bf guided waves}. 

Guided waves can propagate through different configurations of conductors (and insulators). In general, any such system used to convey electromagnetic wave is called 
a {\bf wave guide}. However, this term is usually associated with a specific configuration - a hollow metalloid pipe. Other common configurations are co-axial
transmission lines, parallel lines, rectangular wave guides and two parallel planes. These systems can be put into two groups: 

\begin{itemize}
\item {\bf Transmission lines:} This is a set of two or more conductors that can define a unique voltage and current along the line, thus it is possible to represent such
in a circuit diagram.

\item {\bf Waveguides:}  These are single enclosed conductors (a hollow pipe) with a given crossectional geometry (square, circular, elliptic, triangular, etc...). 
A unique voltage or current along these type of wave guides cannot be specified.
\end{itemize}

For a specific configuration, only certain patterns for electric and magnetic field will satisfy the boundary conditions problem of the differential 
equations that govern the wave's dynamics. These patterns are normally called {\it modes}.
In that case, waves can have components of their electric or magnetic field along the direction of propagation, 
and are conventionally named accordingly:

\begin{itemize}
\item {\bf Transverse Electric (TE):} The magnetic field has a component along the direction of propagation (i.e., a longitudinal component),
but the electric field is entirely in the transverse plane.
\item {\bf Transverse Magnetic (TM):} The electric field has a longitudinal component, but the magnetic field is entirely transverse.
\item {\bf Transverse Electromagnetic (TEM):} Both electric and magnetic fields are contained entirely in the transverse plane.
\item {\bf Hibryd modes (EH and HE):} Both electric and magnetic fields have a longitudinal component.
\end{itemize}
The TEM modes cannot exist in waveguides, but, on the other hand, are the dominant modes in transmission lines. Plane waves meet the TEM criteria, hence
they can be understood as TEM modes as well. Additionally, TEM modes can be excited on any frequency, while TE and TM modes cannot propagate below a 
characteristic \textit{cutoff frequency}.

The electric and magnetic fields propagating through any guided structure, must obey Maxwell's equations, with appropriate boundary conditions.
Let the direction of propagation be aligned with the $z$-axis. The fields can be written as a sum of the transverse component (subscript $t$)
and a longitudinal component (subscript $z$):
$$ \vec E(x,y,z,t) = \vec E_0(x,y,t) e^{-i(\beta z-\omega t)}= (\vec E_t(x,y) + E_z(x,y)\hat e_z)e^{-i(\beta z - \omega t)}$$
$$ \vec B(x,y,z,t) = \vec B_0(x,y,t) e^{-i(\beta z-\omega t)}= (\vec B_t(x,y) + B_z(x,y)\hat e_z)e^{-i(\beta z - \omega t)}$$

In the equations above, the parameter $\beta$ is called the \textit{propagation constant} of the wave. 

In the propagation region, the source free Maxwell's equations are:
$$\nabla \times \vec E = - \mu_0 \frac{\partial \vec B}{\partial t} = i\omega \mu_0 \vec B$$
$$\nabla \times \vec B = - \epsilon_0 \frac{\partial \vec E}{\partial t} = - i\omega \epsilon_0 \vec E$$

The cutoff frequency mentioned before is defined in terms of the respective wave number as $\omega^2_c/c = k^2_c = k^2 - \beta^2$, with $k^2 = \omega / c = \omega \sqrt{\mu_0 \epsilon_0}$

Regarding the boundary conditions, they follow the usual criteria for conductors: The tangential component electric field and the normal component of the 
magnetic field should be zero at the interface with the conducting plates.

\subsection{About the parallel plate transmission line}

Two parallel conducting plates of width $w$, separated by a distance $d$ form what is known as a parallel plate transmission line. 
The space between the plates is filled with some dielectric material, or, simply free space. If the dielectric and the conductors are taken to be perfect,
this yields the so-called "lossless configuration". Other common simplifications include: taking $w \ll d$; taking the plates infinitely long (so reflections
can be neglected); and assuming that the fields have an harmonic time dependence.

This specific configuration, although taken as ideal, is the basis of many practical transmission lines, including the microstrip which has boosted the electronics industry
over the past decades. This kind of transmission line will be the focus of the following sections.

\underline{Note:} In what follows, $c=\frac{1}{\sqrt{\mu_o \epsilon_0}} = 1$

\subsubsection{TEM modes}
For the TEM modes, the electric and magnetic field are only present in the $x$-$y$ plane.
Since $E_z = B_z = 0$ in TEM modes, then from the curl equations of the fields, it follows that the transverse fields would be equally zero.
That is to say there is no wave at all.
The only way to avoid this is to take $k_c = 0$, which gives then indeterminancy of $\frac{0}{0}$. In this case $k_c^2 = 0 \Rightarrow \beta = k$ and in the other,
$k_c^2 = c\omega_c = 0 \Rightarrow \omega_c =0$. In other words, all novanishing frequencies are allowed to propagate in the TEM mode.


The electric then takes the form
$$\vec E(x,y,t) = \vec E_0(x,y,t) e^{-i(kz-\omega t)}
$$
Using this in the wave equation gives:
$$(\nabla^2 - \frac{\partial}{\partial t^2})\vec E = 0 $$
$$(\frac{\partial}{\partial x^2}+\frac{\partial}{\partial y^2})\vec E_0 + (\omega^2-k^2)\vec E_0 = 0 $$

If the electric field oscillates only in the transverse plane, then this configuration is very similar to the one of the parallel plate capacitor's. The electrostatic fields 
(which occur if the wave propagation is frozen in time, for instance at $t=0$) can then be calculated from the well known capacitor equations:
$$\vec E_0 = -\frac{V_0}{d} \hat e_y$$
in which $V_0$ is the potential difference between the plates. The electric and magnetic fields along the plate are then given by:
$$\vec E(x,y,t) =  -\frac{V_0}{d} e^{-ikz}\hat e_y$$
$$\vec B(x,y,t) = \hat e_z \times \vec E = \frac{V_0}{d} e^{-ikz}\hat e_x$$





  \begin{figure*}
  \centerline{
    \mbox{\includegraphics[width=5.00in]{images/plate2.png}}
  }
  \caption{Electric and magnetic field along a parallel plate transmission line.}
  \end{figure*}


It is worth noting that $\vec E$ and $\vec B$ do not depend on $x$ or $y$, hence the fields are uniform  in a cross sectional area. These uniform fields
are associated with induced surface charges and currents.

  \begin{figure*}
  \centerline{
    \mbox{\includegraphics[height=4.00in]{images/plate1.png}}
  }
  \caption{Induced charges and currents on a parallel plate transmission line.}
  \end{figure*}
  
  
%\subsubsection{TE modes}
%\subsubsection{TM modes}

\subsection{Circuit properties of a TEM mode transmission line}

Let $\rho$ be the charge per unit length on the lower plate and $V$ the potential difference between the two plates. The capacitance, $C$, is simply the same one as
a parallel plate capacitor:
$$C=\frac{\rho}{V}=\frac{\epsilon_0 w}{d}$$

From the boundary conditions of the tangential component of the magentic field, the total current density is $ J =  B w$, and the magnetic flux per
unit length is $\Phi = \mu_0 B d$. Hence the inductance per unit length, L, is given by $L=\frac{\Phi}{J}=\frac{\mu_0 d}{w}$. The last paramenter left to calculate
is the characteristic impedance, $Z$, which is related to both impedance and capacitance, according to $Z=\sqrt{\frac{L}{C}}=\frac{d}{w}\sqrt{\frac{\mu_0}{\epsilon_0}}$
With the knowledge of the values $C$, $L$ and $Z$, it is straightforward to come up with the equivalent electronic circuit. The two planes provide the media for current to flow, hence the 
impedance and inductance should be in series with one of the conductors. The capacitance is, by construction, taken in parallel with the conductors. See figure XXX.
  
\begin{figure*}
  \centerline{
    \mbox{\includegraphics[width=2.00in]{images/rlc.png}}
  }
  \caption{Transmission line equivalente circuit.}
  \end{figure*}

\subsection{References}
\begin{itemize}
\item J. D. Jackson - "Classical Electrodynamics", third edition.
\item D. J. Griffiths - "Introduction to Electrodynamics", third edition.
\item An old pdf from unkown author I had from many years ago. If I recal those notes were from MIT.
\end{itemize}

